(3)

Efficiency of Carnot engine is

$\eta =1-\frac{T_2}{T_1}$

Where, $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.

Given that, $\eta =50\%,$ $T_1=327°C=600$ $K$

         $T_2=?$

$\therefore$   $\frac{50}{100}=1-\frac{T_2}{600}\Rightarrow T_2=300 K$

or   $T_2=300-273=27°C$

(1)

The relation between coefficient of performance and efficiency of Carnot engine is given as

       $\beta =\frac{1-\eta }{\eta }.$                    Given $\eta =\frac{1}{10},$ $W=10$ $\text{J}$

        $\beta =\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{{\displaystyle 9}}{{\displaystyle 10}}.10=9$

Since, $\beta =\frac{Q_2}{W},\text{ where }{Q}_2\text{ is the amount of energy absorbed from the reservoir}$

$\therefore$ $Q_2=\beta W=9\times 10=90$$\text{J}$